Replacing for loop in the function

Can someone please help me to remove this for loop in the function below and still get the same result that I get with loops?

Thanks alot in advance.

function I = func(f, a, b, n)
    % Add description, name, date, inputs, outputs
    h = (b-a)/n;
    I = 0;
    for j = 1:n
        x = a + (j-0.5)*h;
        I = I + h*f(x);
    end
end

Answer 1

[Star Strider]

See if ‘func2’ (without the loop) does what you want.

f = @(x) exp(-0.1*x) .* sin(2*pi*x);
a = -5; 
b =  5;
n = 10;

I = func(f,a,b,n)

I = -2.0704e-15

I = func2(f, a, b, n)

I = -2.0704e-15

function I = func(f, a, b, n)
    % Add description, name, date, inputs, outputs
    h = (b-a)/n;
    I = 0;
    for j = 1:n
        x = a + (j-0.5)*h;
        I = I + h*f(x);
    end
end

function I = func2(f, a, b, n)
    h = (b-a)/n;
    j = 1:n;
    x = a + (j-0.5)*h;
    I = sum(h*f(x));
end

Experiment to get different results.

.

[Haider Ali]

thanks alot. it worked perfect. you are a life saviour

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

Answer 2

[Rik]

What you want is not possible. Depending on what f is there might be ways to avoid the loop and use vector-operations instead, but a general solution is not possible.

It is a misconception that loops are always slower. They can be faster, especially when comparing to cellfun or arrayfun, which just hide the loop. The only situation where a loop is faster, is when there is a direct function. Instead of looping through a vector and using +, you can use sum. Or instead of nested loops you might be able to use conv.

For your case there isn't a general solution, because f is unknown, and the variables resulting in x are not guaranteed to result in a scalar.

Answer 3

[Jan]

Does f() accept a row vector as input? Does it reply a row vector or matrix then? If so:

function I = func(f, a, b, n)
h = (b - a) / n;
x = a + (0.5:(n - 0.5)) * h;
I = h * sum(f(x), 2);
end

[Haider Ali]

Thanks alot